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How to find the value of $e^{-x} x^n$ at x = $\infty$ ? Actually while proving that $\Gamma(n+1) = n\Gamma(n)$ there is a step where I need to evaluate ($e^{-x} x^n$)$\big|_0^\infty$ . Now, since this is not an improper integral but the expression after calculating an improper integral(by parts) for which I need to find the value by substituting upper and lower limit, I donot know how to calculate the value at infinity since infinity is not a finite value. Thanks for any help.

Esha
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    This is just a standard limit, which can be evaluated using L'Hopitals rule or using this – Casimir Rönnlöf Sep 10 '20 at 05:26
  • Are you aware that $O(e^x)>O(x^n)$ as $x\to\infty$ for all $n$? –  Sep 10 '20 at 05:28
  • Regarding your other question, which is legitimate: you don't calculate the value at $\infty$, but the limit at $\infty$. To see why, do the integration by parts on $[0,A]$ and after that let $A\to\infty$. – Jean-Claude Arbaut Sep 10 '20 at 05:29
  • @epiliam What is $O$ here ? – Esha Sep 10 '20 at 05:37
  • It's https://en.wikipedia.org/wiki/Big_O_notation. It's how fast it grows as $x\to\infty$. For example $O(3x) = O(\sqrt{x^2-3})$. My point was that $e^x\to\infty$ faster than any polynomial so we must have that $\lim_{x\to\infty}e^{-x} x^n=0$. –  Sep 10 '20 at 05:41

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There is a result that $\lim_{x \rightarrow \infty}\frac{p(x)}{e^x} = 0$ for any polynomial $p(x)$. You can see this intuitively looking at Taylor's Series expansion of $\exp(x)$.

Then, $$\frac{x^n}{e^x} |_0^\infty = \lim_{\alpha \rightarrow \infty} \frac{\alpha^n}{e^\alpha} - \frac{0}{1} = 0. $$

Otherwise, you could use L'Hopital since numerator and denominator are going to $\infty$ and are continuous and differentiable in all $\mathbb R$.

thewatcher
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Exponential growth is faster than polynomial. L'hopital's rule gives $0/e^x\to0$, when applied repeatedly.